Have you ever seen this formula?

What about this one?These are *combinatorial formulas*, used to solve counting problems, and if you’ve been preparing for the GRE, you might be familiar with them as the formulas for *permutations *and *combinations*, two of the most misunderstood and most fun concepts tested in the Quantitative Reasoning section.

## Is There an Easier Way?

The first formula is used to determine how many unique **arrangements** of *k *items you can make when selected from a set containing *n *total items without replacing the items selected. This is the *k*-permutation formula, or simply the **permutation formula **for our purposes.

The second formula is used to determine how many unique **groups** of *k *items you can make when selected from a set containing *n *total items without replacing the items selected. This is the **combination formula**.

Some students are rightly perplexed about how to apply these formulas to actual GRE problems, and their difficulty often is the result of overly abstruse explanations that rely heavily on plugging the correct numbers into one of these formulas without actually understanding how these situations work.

Fortunately, there is a far easier and more transparent way to deal with any problem that arises on the GRE that involves counting. Let’s consider some hypothetical scenarios:

- Jamie is at the make-your-own burrito shop and is choosing his options. He can select either a whole wheat or white tortilla; either chicken, carnitas, or steak; and either red sauce or green sauce. Assuming that he can only select one item from each group, how many possible combinations of ingredients could Jamie choose to make his burrito?
- Casa Linda Movie Theater is making its program for the Thursday evening double feature. The theater manager must select two movies to play one after another. The theater currently has five movies from which she may choose. How many possible schedules could the theater manager make?
- The university graduate students association is planning a conference and must select participants for a four-person panel. Eight papers have been submitted for the association’s review. Among these eight submissions, the association must choose the four authors who will participate on the panel. How many possible discrete panels could the association choose?

**Slots!**

If you’ve got a great memory and have had some practice with the formulas above, you might be able to recognize what goes where and how to plug in your values. However, if you’re like me, you might prefer an approach that is slightly less formula-intensive. Here is another process you can use. Imagine you have a certain number of slots to fill with different possibilities.

- How many slots do you have to fill?
- Are you filling in your slots with selections from the same group of items or from different groups of items?
- Within each group of slots selected from the same group of items, can you reuse items or do you take an item out each time one is selected?
- Within each group of slots selected from the same group of items, does order matter or does order not matter?

Let’s see how this works by solving the hypothetical scenarios outlined above.

**Example Problem 1**

Jamie is at the make-your-own burrito shop and is choosing his options. He can select either a whole wheat or white tortilla; either chicken, carnitas, or steak; and either red sauce or green sauce. Assuming that he can only select one item from each group, how many possible combinations of ingredients could Jamie choose to make his burrito?

- How many slots do you have to fill?

We have 3 slots.

- Are you filling in your slots with selections from the same group of items or from different groups of items?

Different groups.

In this case, we have three different groups and only one selection to make from each group. Therefore we just fill in the possibilities for each slot and multiply the three numbers together to determine the possible combinations.

The answer is 12 possible burritos. So why does this work? Consider this flow-chart illustration:

Follow each path on the flowchart to arrive at a different possibility. Also, remember, always go with the green sauce.

**Example Problem 2**

Casa Linda Movie Theater is making its program for the Thursday evening double feature. The theater manager must select two movies to play one after another. The theater currently has five movies from which she may choose. How many possible schedules could the theater manager make?

- How many slots do you have to fill?

We have 2 slots.

- Are you filling in your slots with selections from the same group of items or from different groups of items?

Same group.

- Within each group of slots selected from the same group of items, can you reuse items or do you take an item out each time one is selected?

Take one out each time.

- Within each group of slots selected from the same group of items, does order matter or does order not matter?

Order matters.

In this case, you start with the slots as above. Then fill each slot with the number of possibilities available for each slot. Remember, since we’re taking one movie away each time, you have to subtract one for each subsequent slot (because there is one fewer possibility each time). We end up with:

So we end up with 20 possible movie schedules.

**Example Problem 3**

The university graduate students association is planning a conference and must select participants for a four-person panel. Eight papers have been submitted for the association’s review. Among these eight submissions, the association must choose the four authors who will participate on the panel. How many possible discrete panels could the association choose?

- How many slots do you have to fill?

We have 4 slots.

Same group.

- Within each group of slots selected from the same group of items, can you reuse items or do you take an item out each time one is selected?

Take one out each time.

- Within each group of slots selected from the same group of items, does order matter or does order not matter?

Order **does not **matter.

In this case if a panel consists of Amy, Bruce, Chava, and Dev or Bruce, Dev, Chava, and Amy, we still have the same panel. If we were just to plug in our numbers as above:

We would be counting the same groups multiple times. We don’t care about the order. We need to reduce the total number to reflect only the unique groups. How do we do this? Well, consider the number of possible arrangements for a group of 4 items:

There are 24 different arrangements of 4 items. You’ve got ABCD, ABDC, ACBD, ACDB, etc. — 24 different arrangements in total! We need to count these 24 arrangements only once. How do you get from 24 to 1? You divide 24 by 24. This is what you do with your slots above. You divide the top group by the bottom group (the bottom group is the factorial of the number of slots — 4!). Your work should look like this:

Notice you can easily reduce to make the problem easier. 3 and 2 combined cancel 6. 4 reduces 8 to 2. You end up with 2 times 7 times 5. The answer is 70 possible unique panels.

**Conclusion**

As with many GRE Quantitative problems, the most difficult part of solving counting problems is setting them up to begin with. However, if you stick to the step-by-step process above, you may find that you are able to follow what’s going on more easily than when you try to plug numbers into the formulas. I hope you also perceive that all that we’ve done with our slots is to create visual illustrations of what is happening in the formulas I gave you at the beginning of this post. **These are the formulas in action! **Now you know what they are actually doing.

On the most difficult GRE problems, counting may be combined with other concepts, such as probability, to create some real challenges; we’ll discuss that in a future post. In the meantime, I hope you find this slot technique helpful.

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